[TenTec] 75 Ohm twin velocity factor ?
Steve Hunt
steve at karinya.net
Sat Jan 27 15:17:34 EST 2007
Jim,
Apologies - my mistake. I misread how the "brackets" were grouped :)
Steve G3TXQ
----- Original Message -----
From: Jim FitzSimons
To: 'Discussion of Ten-Tec Equipment'
Sent: Saturday, January 27, 2007 5:47 PM
Subject: Re: [TenTec] 75 Ohm twin velocity factor ?
This is not a simplification. This is exact. It works for all
values of a/b.
120*ACOSH(b/a)=120*(2*LN(SQRT(b/a-1)+SQRT(b/a+1))-LN(2))
Here is a table using this formula.
b/a, impedance
74.20994852,600
21.27229987,450
6.132289479,300
1.201753692,75
1.185465218,72
1.088068713,50
1.013921068,20
1.003474232,10
1.000034722,1
Jim FitzSimons W7ANF
-----Original Message-----
From: tentec-bounces at contesting.com [mailto:tentec-bounces at contesting.com]
On Behalf Of Steve Hunt
Sent: Saturday, January 27, 2007 6:43 AM
To: Discussion of Ten-Tec Equipment
Subject: Re: [TenTec] 75 Ohm twin velocity factor ?
Jim,
Unless I'm doing something wrong, this simplification fails as b tends to a.
It produces an answer of 83 for a=b so I guess it's the same simplification
as the ARRL's log formula. According to Jerry the simplified formula is
increasingly inaccurate for results below 200.
Thanks all the same.
Steve G3TXQ
----- Original Message -----
From: Jim FitzSimons
To: 'Discussion of Ten-Tec Equipment'
Sent: Saturday, January 27, 2007 10:36 AM
Subject: Re: [TenTec] 75 Ohm twin velocity factor ?
characteristic impedance is 120 cosh^-1 b/a
This is a quote from the help for DERIVE which is a product of TI.
"ACOSH(z) is the inverse hyperbolic cosine of z.
ACOSH(z) simplifies to 2*LN(SQRT(z - 1) + SQRT(z + 1)) - LN(2)"
120*ACOSH(b/a)=120*(2*LN(SQRT(b/a-1)+SQRT(b/a+1))-LN(2))
This is not very simple, but you can calculate the impedance
on most calculators using this formula.
Here is the reverse formula to calculate b/a from impedance z.
b/a=e^(z/120)/2+e^(-z/120)/2
Here is a link to DERIVE http://www.derive.com
Jim FitzSimons W7ANF
-----Original Message-----
From: tentec-bounces at contesting.com [mailto:tentec-bounces at contesting.com]
On Behalf Of Dr. Gerald N. Johnson
Sent: Friday, January 26, 2007 3:26 PM
To: tentec at contesting.com
Subject: Re: [TenTec] 75 Ohm twin velocity factor ?
On Fri, 2007-01-26 at 21:35 +0000, Steve Hunt wrote:
> Folks,
>
> Thanks for all the responses on the Velocity Factor issue.
>
> I would expect 75 Ohm twin to have a lower VF than 300 Ohm or 450 Ohm
line. With nothing but air between the conductors the limit on
characteristic impedance is 83 Ohms, so to achieve 75 Ohm the line must
have
a significant amount of dielectric material as a separator; this will tend
to lower the VF.
That 83 ohm limit is untrue. Its where the conductors would have to
overlap
if you use the log formula which is only accurate above 200 ohms. The
proper
formula for all impedances and spacings that works down to .01 ohm
characteristic impedance is 120 cosh^-1 b/a (that's the inverse
hyperbolic
cosine, not often in a calculator or set of tables) as I recall. That
shows
curved lines on a log log chart.
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